Let $k$ be an algebraically closed field. For $f \in k[x_1,...,x_n]$, write $V(f) \subset k^n$ for the hypersurface defined by $f=0$.
As an application of the NSS, I want to prove the following.
If $f$ is irreducible and $f$ does not divide $g$, then $V(f) \not\subset V(g)$.
My idea is to reason by contradiction and to suppose that $V(f) \subset V(g)$.
Now I consider the ideal generated by $f$. Since $f$ is irreducible, $(f)$ is prime and thus $(f) \neq k[x_1,...,x_n]$.
Since I supposed that $V(f) \subset V(g)$, the NSS (which I can apply since $k$ is algebraically closed) tells me that there exists $n \in \mathbb{N}$ such that $g^n \in (f)$. But this is a contradiction to the fact that $f$ does not divide $g$.
I am really unsure about my reasoning and my understanding of the NSS.
Any comments would be appreciated.
It seems reasonable to start with a contradiction, i.e., that $V(f)\subseteq V(g)$. At this point, however, it may be better to observe that $I(V(f))\supseteq I(V(g))$. You are essentially making such an observation, but the ideal $I(V(f))$ should be identified directly.
Now, the Nullstellensatz (and algebraic closure) comes in to show that $I(V(f))=\langle f\rangle$. This requires the Nullstellensatz and algebraic closure since otherwise $V(f)$ might be empty, so $I(V(f))$ would be larger than $\langle f\rangle$.
Now, if $I(V(f))\supseteq I(V(g))$, then $g\in I(V(f))=\langle f\rangle$. This, however, contradicts the fact that $f$ does not divide $g$.